Question

4.3 - hypothesis tests for the mean
score: 7.5/90 answered: 2/11
question 3
a shareholders group is lodging a protest against your company. the shareholders group claimed that the mean tenure for a chief executive office (ceo) was different than 9 years. a survey of 51 companies reported in the wall street journal found a sample mean tenure of 10.3 years for ceos with a standard deviation of s = 5.6 years (the wall street journal, january 2, 2007). you dont know the population standard deviation but can assume it is normally distributed.
you want to formulate and test the hypothesis made by the group, at a significance level of α = 0.001. your hypotheses are:
$h_0:mu = 9$
$h_a:mu
eq9$

Explanation:

Step1: Calculate the test - statistic

The formula for the t - test statistic in a one - sample t - test is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu$ is the hypothesized population mean, $s$ is the sample standard deviation, and $n$ is the sample size.
Given $\bar{x} = 10.3$, $\mu = 9$, $s = 5.6$, and $n = 51$.
$t=\frac{10.3 - 9}{5.6/\sqrt{51}}$
$=\frac{1.3}{5.6/\sqrt{51}}\approx\frac{1.3}{5.6/7.1414}\approx\frac{1.3}{0.7842}\approx1.66$.

Step2: Determine the degrees of freedom

The degrees of freedom for a one - sample t - test is $df=n - 1$.
So, $df=51 - 1=50$.

Step3: Find the critical values

For a two - tailed test with $\alpha = 0.001$ and $df = 50$, we look up the t - values in the t - distribution table.
The critical values are $t_{\alpha/2,df}=t_{0.0005,50}\approx\pm 3.496$.

Step4: Make a decision

Since the calculated t - value ($t\approx1.66$) is between $- 3.496$ and $3.496$, we fail to reject the null hypothesis $H_0$.

Answer:

We fail to reject the null hypothesis $H_0:\mu = 9$.